1231 - Coin Change (I) 多重背包--背包计数

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最新推荐文章于 2024-06-28 18:29:00 发布

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本文介绍了一个基于多重背包思想的问题：在一个特殊的商店里，有n种不同面值的硬币，每种硬币的数量有限，求组合成特定金额的方法数量。通过示例详细解释了输入输出格式，并提供了一个C++实现代码。

In a strange shop there are n types of coins of value A1, A2 ... An. C1, C2, ... Cn denote the number of coins of value A1, A2 ... An respectively. You have to find the number of ways you can make K using the coins.

For example, suppose there are three coins 1, 2, 5 and we can use coin 1 at most 3 times, coin 2 at most 2 times and coin 5 at most 1 time. Then if K = 5 the possible ways are:

1112

122

So, 5 can be made in 3 ways.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 50) and K (1 ≤ K ≤ 1000). The next line contains 2n integers, denoting A1, A2 ... An, C1, C2 ... Cn (1 ≤ Ai ≤ 100, 1 ≤ Ci ≤ 20). All Ai will be distinct.

Output

For each case, print the case number and the number of ways K can be made. Result can be large, so, print the result modulo 100000007.

Sample Input

Output for Sample Input

3 5

1 2 5 3 2 1

4 20

1 2 3 4 8 4 2 1

Case 1: 3

Case 2: 9

PROBLEM SETTER: JANE ALAM JAN

利用多重背包思想。

#include<bits/stdc++.h>
# define ll long long
using namespace std;
const ll mod=100000007;
const int N=1e4;
int num[N];
int a[N];//价值
ll dp[N];
int main()
{
    int t;
    cin>>t;
    for(int o=1;o<=t;o++)
    {
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)cin>>a[i];
        for(int i=1;i<=n;i++)cin>>num[i];
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=1;i<=n;i++)
        {
            //n件
            for(int j=m;j>=a[i];j--)
            {
                for(int k=0;k<=num[i];k++)//0-num[i]件件数
                {
                    if(j-k*a[i]<0)break;
                    if(k==0)continue;
                    dp[j]=dp[j]+dp[j-k*a[i]]%mod;//取或不取

                }
            }
        }
        printf("Case %d: %d\n",o,dp[m]%mod);
    }
}